Find the equation of motion of a force of a spring

Question

A spring with a spring constant k of 20 pounds per foot is loaded with a 10-pound weight and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight experiences a retarding force in pounds equal to four times the velocity at every point, find the equation of motion.

Answer 1

Let

k = 20 lb/ft be the spring constant,

m = 10 lb be the weight,

A = -1 ft be the initial displacement,

r = 4 be the coefficient of the retarding force,

g be gravitational acceleration (whose value we will actually not need),

x(t) be the displacement of the weight below the equilibrium position at time t,

v(t) be the velocity of the weight at time t,

a(t) be the acceleration of the weight at time t,

F_g = mg be the (constant) force of gravity, directed downward,

F_r(t) be the force of resistance, directed opposite to velocity,

F_s(t) be the force of the spring, directed opposite to displacement,

F(t) = F_g + F_r(t) + F_s(t) be the net force acting on the weight at time t.

Lets consider downward direction as negative and upward direction as positive.

That is why the initial displacement A is negative (and so is the value of g).

Actually any particular choice of positive direction is irrelevant as long as all you

want is to derive the equation of motion.

It would become relevant only if we wanted a particular solution.

For readability I am dropping the parameter t from x(t), v(t), a(t), ...

After the weight is placed on the spring, it compresses the string to the equilibrium position,

which is some distance E below the uncompressed position.

We now calculate E.

In that equilibrium position

F_s = -kE by the spring equation

F_r = 0 because velocity is 0

F = 0 because acceleration is 0

Thus

0 = F = F_g + F_r + F_s = mg + 0 - kE

E = mg/k

Later when the weight is oscillating we have the following equations:

F_s = -k(x + E) by the spring equation

F_r = -rv according to specification

F = ma by Newton's second law

This can be rewritten as

F_g + F_s + F_r = ma

mg + -k(x + mg/k) - rv = ma

ma + rv + kx = 0

Using the definitions

v = dx/dt

a = d²x/dt²

we get the differential equation

md²x/dt² + rdx/dt + kx = 0

The initial displacement of 1 ft does not enter into the equation,

as that would be used only to determine a specific solution.

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As you suggested, the general solution to the equation of motion is of the form

x = Ae^-t/5sin(7t/5 + B) (1)

for some constants A, B.

I am using the more customary variable t for time instead of y.

Also there two constants to determine -- the amplitude A and the phase B.

We need to calculate the derivative, i.e. speed:

dx/dt = Ae^-t/5(7cos(7t/5 + B) - sin(7t/5 + B)) / 5 (2)

The constants A, B are determined from initial conditions:

x(0) = 1 (initial displacement is 1 ft)

dx/dt(0) = 0 (initial speed is 0)

Substituting t=0 into (1) and (2)

Asin(B) = 1 (3)

A(7cos(B) - sin(B)) / 5 = 0 (4)

Equation (4) can be simplified into

7cos(B) - sin(B) = 0

tan(B) = 7

B ≈ 0.45π

After substituting B into equation (3)

A ≈ 1.01

So the specific solution to the equation of motion is

x = 1.01 e^-t/5sin(7t/5 + 0.45π)